Function Transformations - IB Mathematics Analysis and Approaches

The graph of $y = e^x$ is reflected in the line $y=2$ . What is the equation of the transformed graph?
Answer: $y = 4 - e^x$ A reflection of a point $(x, y)$ across the horizontal line $y=c$ results in a new point $(x, 2c-y)$ . In this case, $c=2$ . The transformation is $(x, y) \to (x, 4-y)$ . Let the new coordinates be $(x', y')$ . So, $x' = x$ and $y' = 4-y$ . We need to express the original coordinates in terms of the new ones: $x = x'$ and $y = 4-y'$ . Substitute these into the original equation $y = e^x$ : $4 - y' = e^{x'}$ Now, solve for $y'$ : $y' = 4 - e^{x'}$ Dropping the primes, the equation of the transformed graph is $y = 4 - e^x$ .
The function $y=f(x)$ is transformed to $y=af(x-b)$ . The point $(2,3)$ on the graph of $f(x)$ is mapped to the point $(5,12)$ on the transformed graph. Find the values of $a$ and $b$ .
Answer: $a=4, b=3$ When you encounter function transformations, you need to understand how parameters affect the graph. The transformation $y = af(x-b)$ involves two changes: $a$ creates a vertical stretch/compression, and $b$ creates a horizontal shift. To find the parameters, use the fact that if point $(p,q)$ is on the original function $f(x)$ , then point $(p+b, aq)$ will be on the transformed function $y = af(x-b)$ . This is because when $x = p+b$ , we get $y = af((p+b)-b) = af(p) = aq$ . Given that $(2,3)$ on $f(x)$ maps to $(5,12)$ on the transformed graph, we can set up equations: - For the x-coordinate: $2 + b = 5$ , so $b = 3$ - For the y-coordinate: $3a = 12$ , so $a = 4$ Let's check each option: **A) $a=4, b=-3$ **: This would map $(2,3)$ to $(2+(-3), 4 \cdot 3) = (-1,12)$ , not $(5,12)$ . **B) $a=4, b=3$ **: This maps $(2,3)$ to $(2+3, 4 \cdot 3) = (5,12)$ ✓ **C) $a=\frac{1}{4}, b=3$ **: This would map $(2,3)$ to $(5, \frac{1}{4} \cdot 3) = (5, \frac{3}{4})$ , not $(5,12)$ . **D) $a=2, b=4$ **: This would map $(2,3)$ to $(6,6)$ , not $(5,12)$ . Study tip: Remember the transformation pattern: horizontal shifts add to the x-coordinate of original points, while vertical stretches multiply the y-coordinate. Always verify your answer by checking if the given point mapping works with your parameters.
The function $f(x) = \frac{x+1}{x-3}$ has a vertical asymptote at $x=3$ and a horizontal asymptote at $y=1$ . Find the equations of the asymptotes of the function $g(x) = -f(2x) + 4$ .
Answer: $x = \frac{3}{2}, y = 3$ The transformations from $f(x)$ to $g(x)$ are: 1. A horizontal stretch by a factor of $\frac{1}{2}$ (from $2x$ ). 2. A reflection in the x-axis (from the negative sign before $f$ ). 3. A vertical translation up by 4 units. The vertical asymptote $x=3$ is only affected by horizontal transformations. Applying the horizontal stretch by $\frac{1}{2}$ , the new vertical asymptote is at $x = \frac{3}{2}$ . The horizontal asymptote $y=1$ is only affected by vertical transformations. Applying the reflection in the x-axis, $y=1$ becomes $y=-1$ . Then, applying the vertical translation up by 4, $y=-1$ becomes $y = -1+4 = 3$ . Thus, the new asymptotes are $x = \frac{3}{2}$ and $y=3$ .
The point $P(4, -6)$ lies on the graph of $y=f(x)$ . The graph is transformed to $y = -f(\frac{1}{2}x+1) + 3$ . Find the coordinates of the image of point P.
Answer: $(6, 9)$ Let the original point be $(x_0, y_0) = (4, -6)$ and the new point be $(x_{new}, y_{new})$ . The horizontal transformation is found by setting the argument of the new function equal to the original x-coordinate: $\frac{1}{2}x_{new}+1 = x_0 = 4$ . Solving for $x_{new}$ : $\frac{1}{2}x_{new} = 3 \implies x_{new} = 6$ . The vertical transformation is found by applying the outer transformations to the original y-coordinate: $y_{new} = -y_0 + 3$ . Substituting $y_0 = -6$ : $y_{new} = -(-6) + 3 = 6 + 3 = 9$ . The coordinates of the image of point P are $(6, 9)$ .
The graph of the function $y = \ln(x)$ is translated 3 units to the left and then reflected in the y-axis. What is the equation of the resulting graph?
Answer: $y = \ln(3-x)$ We apply the transformations sequentially to $y = \ln(x)$ . 1. Translate 3 units to the left: This replaces $x$ with $x+3$ . The function becomes $y = \ln(x+3)$ . 2. Reflect in the y-axis: This replaces $x$ with $-x$ in the current function. The function becomes $y = \ln(-x+3)$ , which can be written as $y = \ln(3-x)$ . Distractor A, $y = \ln(-x-3)$ , results from applying the reflection first and then the translation. If we start with $y=\ln(x)$ , reflection gives $y=\ln(-x)$ , and translation left gives $y=\ln(-(x+3)) = \ln(-x-3)$ . The order of operations matters.
The graph of $y=f(x)$ is transformed to the graph of $g(x) = 2f(1-x)$ . The graph of $y=g(x)$ is the line with equation $y = 6x - 4$ . Find the equation of $f(x)$ .
Answer: $f(x) = 1-3x$ We are given that $g(x) = 6x-4$ and $g(x) = 2f(1-x)$ . Therefore, $2f(1-x) = 6x-4$ . First, divide by 2: $f(1-x) = 3x-2$ . To find $f(x)$ , we perform a substitution. Let $u = 1-x$ . We need to express $x$ in terms of $u$ . From $u = 1-x$ , we get $x = 1-u$ . Now substitute $x = 1-u$ into the equation $f(1-x) = 3x-2$ : $f(u) = 3(1-u) - 2$ $f(u) = 3 - 3u - 2$ $f(u) = 1 - 3u$ Since this holds for any $u$ , we can write the function in terms of $x$ as $f(x) = 1-3x$ .
The graph of $y=f(x)$ is transformed to the graph of $y=f(|x|)$ . This transformation has what effect on the graph of $y=f(x)$ ?
Answer: The part of the graph for $x<0$ is replaced by a reflection of the part for $x>0$ in the y-axis. When you encounter transformations involving absolute value functions, you need to carefully analyze what happens to the input versus the output of the function. The transformation from $y = f(x)$ to $y = f(|x|)$ affects the input to the function, not the output. The key insight is understanding what $|x|$ does: it makes all negative x-values positive while leaving positive x-values unchanged. So when $x > 0$ , we have $|x| = x$ , meaning $f(|x|) = f(x)$ — the right side of the graph stays the same. But when $x < 0$ , we have $|x| = -x$ (which is positive), so $f(|x|) = f(-x)$ — we're now using the function values that originally corresponded to positive x-values. This means the part of the original graph for $x > 0$ gets reflected across the y-axis to replace the part for $x < 0$ . The correct answer is C. Looking at the wrong answers: A suggests the entire graph reflects in the y-axis, but only the left side changes while the right side remains identical. B incorrectly focuses on reflecting parts based on y-values rather than x-values — this transformation doesn't involve the x-axis at all. D suggests reflection in the x-axis, which would affect y-values, not the domain transformation we're seeing here. Strategy tip: For absolute value transformations, remember that $f(|x|)$ always creates an even function (symmetric about the y-axis). The right side of the original graph survives, and its reflection replaces the left side.
The equation $f(x)=0$ has roots at $x=2$ and $x=-5$ . What are the roots of the equation $f(2x-1)=0$ ?
Answer: $x=-2, x=\frac{3}{2}$ The roots of $f(x)=0$ are the values of the input to $f$ that make the output zero. For the equation $f(2x-1)=0$ , the output will be zero when the input, $2x-1$ , is equal to the original roots, 2 and -5. We set up two equations: 1. $2x-1 = 2 \implies 2x = 3 \implies x = \frac{3}{2}$ 2. $2x-1 = -5 \implies 2x = -4 \implies x = -2$ The new roots are $x=-2$ and $x=\frac{3}{2}$ . Distractor B is a common error where the transformation $2x-1$ is applied directly to the roots. Distractor D results from a sign error when solving for x.
The graph of a function $f(x)$ is reflected in the y-axis, then stretched vertically by a factor of 3, and finally translated 2 units down. Which of the following equations represents the final transformed function, $g(x)$ ?
Answer: $g(x) = 3f(-x) - 2$ We apply the transformations step-by-step to $y=f(x)$ : 1. Reflection in the y-axis: This transformation replaces $x$ with $-x$ . The function becomes $y = f(-x)$ . 2. Stretched vertically by a factor of 3: This transformation multiplies the entire function by 3. The function becomes $y = 3f(-x)$ . 3. Translated 2 units down: This transformation subtracts 2 from the entire function. The function becomes $y = 3f(-x) - 2$ . Therefore, the final equation is $g(x) = 3f(-x) - 2$ . Distractor B represents translating down before stretching, which would result in $g(x) = 3f(-x) - 6$ .
The graph of $y = f(x)$ is transformed to $y = f(4-2x)$ . Which of the following describes the sequence of horizontal transformations?
Answer: A horizontal stretch by a factor of $\frac{1}{2}$ , a reflection in the y-axis, and a translation of 2 units to the right. When analyzing horizontal transformations of functions, you need to work systematically through the transformation $y = f(4-2x)$ by rewriting it in standard form and identifying each component. First, factor out the coefficient of $x$ : $y = f(4-2x) = f(-2(x-2))$ . This reveals the transformation structure more clearly. Now you can identify three distinct horizontal transformations applied in sequence: 1. Horizontal stretch by factor $\frac{1}{2}$ (from the coefficient $-2$ , where the stretch factor is $\frac{1}{|coefficient|} = \frac{1}{2}$ ) 2. Reflection in the y-axis (from the negative sign in $-2$ ) 3. Translation 2 units to the right (from $x-2$ , which shifts the graph right by 2) This matches option D exactly. Option A incorrectly identifies a translation of 4 units right instead of 2, and misses the reflection entirely. Option B correctly identifies the stretch factor as 2 instead of $\frac{1}{2}$ - this is a common error where students forget that the stretch factor is the reciprocal of the coefficient. It also incorrectly states a 2-unit left translation. Option C makes the same translation error as A, claiming 4 units right instead of 2. Study tip: Always rewrite transformations in the form $f(a(x-h))$ to clearly see: stretch factor = $\frac{1}{|a|}$ , reflection occurs when $a<0$ , and horizontal shift = $h$ (right if positive, left if negative). This systematic approach prevents mixing up the transformation components.
The function $y=f(x)$ undergoes a sequence of transformations to become $g(x) = 2f(x+3) - 5$ . A point on the graph of $y=g(x)$ is $(1, 9)$ . Find the coordinates of the corresponding point on the graph of $y=f(x)$ .
Answer: $(4, 7)$ Let the original point on $y=f(x)$ be $(x_0, y_0)$ . The transformation $g(x) = 2f(x+3) - 5$ maps this point to a new point $(x_{new}, y_{new})$ . The horizontal transformation is $x \to x-3$ and the vertical transformation is $y \to 2y-5$ . Therefore, $x_{new} = x_0 - 3$ and $y_{new} = 2y_0 - 5$ . We are given the new point is $(1, 9)$ . So, we need to solve for $(x_0, y_0)$ . For the x-coordinate: $1 = x_0 - 3 \implies x_0 = 4$ . For the y-coordinate: $9 = 2y_0 - 5 \implies 14 = 2y_0 \implies y_0 = 7$ . The corresponding point on the graph of $y=f(x)$ is $(4, 7)$ .
The graph of $y=f(x)$ is reflected in the x-axis, and then translated horizontally by 4 units to the left. The resulting graph is given by the equation $y=g(x)$ . Which equation correctly expresses $f(x)$ in terms of $g(x)$ ?
Answer: $f(x) = -g(x-4)$ Function transformations in IB Mathematics require you to think carefully about the order of operations and how they affect the relationship between functions. When working backwards from a transformed function to find the original, you need to reverse each transformation step by step. Let's trace through what happens to $f(x)$ to become $g(x)$ . First, $f(x)$ is reflected in the x-axis, giving us $-f(x)$ . Then this reflected function is translated 4 units left, which means replacing $x$ with $(x+4)$ , resulting in $g(x) = -f(x+4)$ . To find $f(x)$ in terms of $g(x)$ , we reverse these steps. From $g(x) = -f(x+4)$ , we first undo the reflection by multiplying both sides by $-1$ : $-g(x) = f(x+4)$ . Then we undo the horizontal translation by replacing $x$ with $(x-4)$ : $f(x) = -g(x-4)$ . This confirms answer B is correct. Looking at the wrong answers: A gives $f(x) = -g(x+4)$ , which would represent translating right instead of left when reversing the transformation. C gives $f(x) = g(-x-4)$ , incorrectly introducing a reflection about the y-axis that wasn't in the original problem. D gives $f(x) = -g(x)+4$ , which confuses horizontal translation with vertical translation and gets the algebraic relationship completely wrong. Remember: when reversing transformations, work backwards through each step. Horizontal shifts of "left $a$ " become "right $a$ " when reversed, and reflections are their own inverse operations.
The function $f(x)$ is periodic with period 4. What is the period of the function $g(x) = f(2x - 1)$ ?
Answer: 2 A function $f(x)$ has period $P$ if $f(x+P) = f(x)$ for all $x$ . We are given that $f(x)$ has a period of 4, so $f(x+4) = f(x)$ . We want to find the period $T$ of $g(x) = f(2x-1)$ , which means we need to find the smallest positive $T$ such that $g(x+T) = g(x)$ . $g(x+T) = f(2(x+T) - 1) = f(2x + 2T - 1)$ . We want this to be equal to $g(x) = f(2x-1)$ . So we need $f(2x-1+2T) = f(2x-1)$ . Since the period of $f$ is 4, the expression inside $f$ must differ by a multiple of 4. We want the smallest positive $T$ , so we set the added term equal to 4. $2T = 4$ $T = 2$ . The period of $g(x)$ is 2. Horizontal translations (like the -1) do not affect the period.
The graph of $y=\sqrt{x}$ is transformed into the graph of $y=a\sqrt{x-h}+k$ . The transformed graph passes through the points $(5, 7)$ and $(10, 11)$ and has its domain as $x \ge 1$ . Find the value of $a$ .
Answer: 4 The domain of $y=\sqrt{x}$ is $x \ge 0$ , and its endpoint is at $(0,0)$ . The transformation $y=a\sqrt{x-h}+k$ shifts the endpoint to $(h,k)$ and changes the domain to $x \ge h$ . Given that the domain of the transformed graph is $x \ge 1$ , we have $h=1$ . The function is $y = a\sqrt{x-1}+k$ . Now, substitute the two given points into this equation to form a system of linear equations: For $(5, 7)$ : $7 = a\sqrt{5-1} + k \implies 7 = a\sqrt{4} + k \implies 7 = 2a + k$ . For $(10, 11)$ : $11 = a\sqrt{10-1} + k \implies 11 = a\sqrt{9} + k \implies 11 = 3a + k$ . Subtract the first equation from the second: $(11 - 7) = (3a - 2a) + (k - k)$ which gives $4 = a$ . So the value of $a$ is 4.
The graph of $y=f(x)$ is transformed by a vertical stretch with scale factor 2, followed by a translation by the vector $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ . If the point $(1, 4)$ is on the original graph, what are the coordinates of its image?
Answer: $(4, 7)$ When you encounter function transformations, you need to apply them in the correct order to track how points move. Function transformations follow specific rules: vertical stretches multiply y-coordinates, and translations shift points by adding to the coordinates. Starting with the point $(1, 4)$ on the original graph $y = f(x)$ , let's apply each transformation step by step. First, the vertical stretch with scale factor 2 multiplies the y-coordinate by 2 while leaving the x-coordinate unchanged. So $(1, 4)$ becomes $(1, 8)$ . Next, the translation by vector $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ means we add 3 to the x-coordinate and subtract 1 from the y-coordinate. Applying this to $(1, 8)$ gives us $(1 + 3, 8 - 1) = (4, 7)$ . Looking at the wrong answers: Answer A $(6, 3)$ appears to have confused the transformations, possibly applying the stretch incorrectly or mixing up the order. Answer B $(4, 1)$ gets the x-coordinate right but seems to have subtracted rather than added during the vertical stretch. Answer D $(6, 7)$ correctly handles the y-transformations but incorrectly adds the translation vector's x-component twice, giving $1 + 3 + 2 = 6$ instead of $1 + 3 = 4$ . The correct answer is C $(4, 7)$ . Strategy tip: Always apply transformations in the order given, and remember that vertical stretches affect only y-coordinates while translations affect both coordinates by adding the vector components. Write out each step to avoid mixing up the operations.
The function $y=f(x)$ has domain $[-4, 6]$ and range $[-2, 8]$ . Determine the domain and range of the function $g(x) = \frac{1}{2}f(4-2x) + 1$ .
Answer: Domain: $[-1, 4]$ , Range: $[0, 5]$ For the domain, we consider the argument of the function $f$ . The original domain is $-4 \le x_{\text{old}} \le 6$ . The new function has argument $4-2x$ . So we set $-4 \le 4-2x \le 6$ . Subtract 4 from all parts: $-8 \le -2x \le 2$ . Divide by -2 and reverse the inequalities: $4 \ge x \ge -1$ , which is $-1 \le x \le 4$ . The new domain is $[-1, 4]$ . For the range, we consider the transformations outside the function $f$ . The original range is $-2 \le f(x) \le 8$ . The new function is $g(x) = \frac{1}{2}f(...) + 1$ . Multiply the range by $\frac{1}{2}$ : $\frac{1}{2}(-2) \le \frac{1}{2}f(...) \le \frac{1}{2}(8) \implies -1 \le \frac{1}{2}f(...) \le 4$ . Then add 1: $-1+1 \le \frac{1}{2}f(...) + 1 \le 4+1 \implies 0 \le g(x) \le 5$ . The new range is $[0, 5]$ .
Let $f(x) = x^2$ . The graph of $f(x)$ is translated 2 units to the left and 3 units down to obtain the graph of $g(x)$ . Then, the graph of $g(x)$ is reflected in the x-axis to obtain the graph of $h(x)$ . Find the equation of $h(x)$ .
Answer: $h(x) = -(x+2)^2 + 3$ When you encounter function transformations, work systematically through each transformation step by step, applying them to the original function in the order given. Starting with $f(x) = x^2$ , let's find $g(x)$ first. Translating 2 units left means replacing $x$ with $(x+2)$ , and translating 3 units down means subtracting 3 from the entire function. Therefore: $g(x) = (x+2)^2 - 3$ Next, reflecting $g(x)$ in the x-axis means multiplying the entire function by -1: $h(x) = -g(x) = -[(x+2)^2 - 3] = -(x+2)^2 + 3$ Now let's examine why each answer choice is wrong or right: Choice A, $h(x) = (-x+2)^2 + 3$ , incorrectly applies the horizontal translation. This represents a translation 2 units to the right, not left, and doesn't include the reflection properly. Choice B, $h(x) = -(x-2)^2 - 3$ , has two errors: it translates right instead of left (using $x-2$ instead of $x+2$ ), and it moves down instead of reflecting the vertical translation correctly after the x-axis reflection. Choice C, $h(x) = -(x+2)^2 + 3$ , correctly shows the left translation with $(x+2)$ , the downward translation of the original function, and the x-axis reflection that flips the sign of the entire expression. Choice D, $h(x) = -(x+2)^2 - 3$ , correctly handles the horizontal translation and reflection but fails to account for how the x-axis reflection affects the vertical translation. Study tip: Remember that x-axis reflections flip the sign of the entire function, which affects how vertical translations appear in the final equation.
A function $f(x)$ has $f(3)=5$ . What is the value of $g(1)$ if $g(x) = 2f(x+2) - 3$ ?
Answer: 7 We need to find the value of $g(1)$ . We substitute $x=1$ into the equation for $g(x)$ . $g(1) = 2f(1+2) - 3$ $g(1) = 2f(3) - 3$ We are given that $f(3)=5$ . We can substitute this value into the expression: $g(1) = 2(5) - 3$ $g(1) = 10 - 3 = 7$ . The value of $g(1)$ is 7.
The function $y=f(x)$ is transformed to $y=f(x+c)-c$ . If the graph of the transformed function passes through the origin, which statement must be true about the original function $f(x)$ ?
Answer: It must pass through the point $(c, c)$ . When you encounter function transformations, the key is understanding how changes inside and outside the function notation affect the graph's position. The transformation $y = f(x + c) - c$ combines two shifts: $f(x + c)$ shifts the graph left by $c$ units, and subtracting $c$ shifts it down by $c$ units. If this transformed function passes through the origin $(0, 0)$ , then substituting these coordinates gives us: $0 = f(0 + c) - c = f(c) - c$ , which means $f(c) = c$ . This tells us the original function must pass through the point $(c, c)$ , making answer D correct. Let's examine why the other options fail. Option A claims the y-intercept must be $c$ . The y-intercept occurs where $x = 0$ , so $f(0)$ would need to equal $c$ . However, our condition only tells us that $f(c) = c$ , not $f(0) = c$ . Option B suggests the original function passes through $(-c, -c)$ . This would mean $f(-c) = -c$ , but again, we only know $f(c) = c$ . Option C claims the original function passes through the origin, meaning $f(0) = 0$ . This isn't necessarily true since we established $f(c) = c$ , not $f(0) = 0$ . Remember this pattern: when a transformed function passes through a specific point, work backwards by substituting those coordinates into the transformation equation. This algebraic approach eliminates guesswork and reveals exactly what conditions the original function must satisfy.

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